Question

calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCL k of acetic acid is 1.8×10^-5

Answer 1

First we should write the reaction,
For the HCl...............> H+........+..Cl-
…........0.1M.................0................0
............0....0.1M.0.1M.after its complete dissociation
[Here ahve neglected 10-7M of H+ due to water dissociation because 0.1M is more than 100 times than 10-7M]
Now for the weak acid which not dissociate completely and always form an equilibrium:
CH3COOH+ H2O=CH3COO-+.H+
0.01M.0.0.1M(from dissociation of HCl) .initially
0.01-x..x.x+0.1M.after dissociation of acetic acid
Now writing
Ka=[CH3COO-][H+]/[CH3COOH]
1.8*10-5 = x*(0.1+x)/0.01
Approximation:the acid is weak and soassuming x<<0.1M,0.1+x~0.1 M
0.1x= 1.8*0.01*10-5 M
so x= 1.8*10-6 M
so % onisation of acid=x/0.01 *100=1.8*10-2%=0.018%(very less due to common ion effect)